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PhyHardAdv 2014 • Paper 2

Total Work Performed by Gases with Frictionless Movable Partition

Comprehension Passage

The diagram illustrates an insulated container fitted with a frictionless, movable top piston. All outer walls and the piston consist of perfectly insulating material, preventing thermal exchange with the environment. The inner volume is split into two sections by a stationary, thermally conductive partition, facilitating gradual heat transfer between the chambers. The bottom chamber holds 2 moles2\text{ moles} of a monatomic ideal gas initially at 700 K700\text{ K}, while the upper chamber contains 2 moles2\text{ moles} of a diatomic ideal gas at an initial temperature of 400 K400\text{ K}. Molar heat capacities are given as CV=32R,CP=52RC_V = \frac{3}{2}R, C_P = \frac{5}{2}R for the monatomic gas, and CV=52R,CP=72RC_V = \frac{5}{2}R, C_P = \frac{7}{2}R for the diatomic gas.

Passage Diagram 1

Suppose now that the separating partition is free to slide without friction, maintaining equal pressures across both compartments. Calculate the total work done by the gases up to the moment complete thermal and mechanical equilibrium is attained.

Options

A

250 R250\text{ }R

B

200 R200\text{ }R

C

100 R100\text{ }R

D

−100 R-100\text{ }R

Correct

Topics & Concepts

ThermodynamicsFirst Law of Thermodynamics

Step-by-Step Solution

To find the total work performed by the gases until thermal and mechanical equilibrium is reached, we analyze the system using the First Law of Thermodynamics and heat transfer equations under constant pressure conditions.

1. Condition for Equilibrium and Final Temperature

  • The upper piston is movable without friction under a constant external pressure P0P_0.
  • The internal partition is also free to slide without friction, which ensures that the pressure in both compartments remains equal at all times: P1=P2=P0=constantP_1 = P_2 = P_0 = \text{constant}

Since both gases undergo a process at constant pressure and the overall system is thermally insulated from the environment (Qnet=0Q_{\text{net}} = 0), the heat released by the monatomic gas equals the heat absorbed by the diatomic gas at constant pressure: Q1+Q2=0Q_1 + Q_2 = 0

Using the constant-pressure molar heat capacities CPC_P: n1CP1(Tf−T1,i)+n2CP2(Tf−T2,i)=0n_1 C_{P1} (T_f - T_{1,i}) + n_2 C_{P2} (T_f - T_{2,i}) = 0

Given data:

  • Monatomic gas (n1=2 molesn_1 = 2\text{ moles}): T1,i=700 KT_{1,i} = 700\text{ K}, CP1=52RC_{P1} = \frac{5}{2}R
  • Diatomic gas (n2=2 molesn_2 = 2\text{ moles}): T2,i=400 KT_{2,i} = 400\text{ K}, CP2=72RC_{P2} = \frac{7}{2}R

Substituting these values into the energy balance equation: 2(52R)(Tf−700)+2(72R)(Tf−400)=02 \left(\frac{5}{2}R\right) (T_f - 700) + 2 \left(\frac{7}{2}R\right) (T_f - 400) = 0

5R(Tf−700)+7R(Tf−400)=05R(T_f - 700) + 7R(T_f - 400) = 0

5Tf−3500+7Tf−2800=05 T_f - 3500 + 7 T_f - 2800 = 0

12Tf=6300  ⟹  Tf=525 K12 T_f = 6300 \implies T_f = 525\text{ K}

2. Calculation of Total Work Done

According to the First Law of Thermodynamics applied to the composite system: Qnet=ΔU+WtotalQ_{\text{net}} = \Delta U + W_{\text{total}}

Since the entire system is thermally insulated from the surroundings, Qnet=0Q_{\text{net}} = 0, which gives: Wtotal=−ΔUW_{\text{total}} = -\Delta U

The total change in internal energy ΔU\Delta U is the sum of changes in internal energy of both gases: ΔU=ΔU1+ΔU2=n1CV1(Tf−T1,i)+n2CV2(Tf−T2,i)\Delta U = \Delta U_1 + \Delta U_2 = n_1 C_{V1} (T_f - T_{1,i}) + n_2 C_{V2} (T_f - T_{2,i})

Substituting the given values (CV1=32RC_{V1} = \frac{3}{2}R and CV2=52RC_{V2} = \frac{5}{2}R): ΔU1=2(32R)(525−700)=3R(−175)=−525R\Delta U_1 = 2 \left(\frac{3}{2}R\right) (525 - 700) = 3R (-175) = -525 R

ΔU2=2(52R)(525−400)=5R(125)=625R\Delta U_2 = 2 \left(\frac{5}{2}R\right) (525 - 400) = 5R (125) = 625 R

ΔU=−525R+625R=100R\Delta U = -525 R + 625 R = 100 R

Therefore, the total work done by the gases is: Wtotal=−ΔU=−100RW_{\text{total}} = -\Delta U = -100 R

Correct Option: (D)

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